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Likurg_2 [28]
3 years ago
8

Technician A says that power is the rate that energy is stored. Technician B says that power refers to the rate that energy is t

ransformed into another kind of energy. Who is correct?
Select one:
a. Technician A
b. Technician B
c. Both technicians A and B
d. Neither technician A nor B
Physics
1 answer:
Alex777 [14]3 years ago
7 0

Answer:

C. Both technicians A and B

Explanation:

From the physical definition, power is defined as the rate of a body doing work. It is expressed as  

                                     P = w/t    watts

Where  

                       w - is the work done or the energy of the system in joules

                        t - time

The unit of power is represented in watts.

Whenever there is a rate of change of energy in the system, it accounts for the efficiency of the power of the system.

Hence, the statements of both technicians are correct.

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Answer:

I belive it would be ture

Explanation:

It's been a while since I learned this but I think that is right.

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A person pushes a 10 kg box from rest and accelerates it to a speed of 4 m/s with a constant force. If the box is pushed for a t
lutik1710 [3]

The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration <em>a</em> is such that

4 m/s = <em>a</em> (2.5 s)   →   <em>a</em> = (4 m/s) / (2.5 s) = 1.6 m/s²

Then the force applied to the box has a magnitude <em>F</em> such that

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The rate of turn at any airspeed is dependent upon
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A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?
Olenka [21]
Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
F = k*x

Solving: 
F = k*x
F = 50*0.15
\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
E = \frac{k*x^2}{2}

Solving:(Energy associated with this stretching)
E = \frac{k*x^2}{2}
E =  \frac{50*0.15^2}{2}
E =  \frac{50*0.0225}{2}
E =  \frac{1.125}{2}
\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark

7 0
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In the diagram, q1=+6.25 * 10^ -8 C. What is the potential difference when you go from point A to point B? Include the correct s
Nimfa-mama [501]

Answer:

Moving a unit "positive" test charge from A to B will result in a reduction in potential

V = K Q / R      potential at a point

V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q

V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8

V2 - V1 = -4.17 * 562.5 J/C

V = - 2346 Volts

7 0
3 years ago
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